Problem: You have found the following ages (in years) of all 6 porcupines at your local zoo: $ 10,\enspace 8,\enspace 9,\enspace 16,\enspace 7,\enspace 10$ What is the average age of the porcupines at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 6 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{10 + 8 + 9 + 16 + 7 + 10}{{6}} = {10\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $10$ years $0$ years $0$ years $^2$ $8$ years $-2$ years $4$ years $^2$ $9$ years $-1$ years $1$ year $^2$ $16$ years $6$ years $36$ years $^2$ $7$ years $-3$ years $9$ years $^2$ $10$ years $0$ years $0$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0} + {4} + {1} + {36} + {9} + {0}} {{6}} $ $ {\sigma^2} = \dfrac{{50}}{{6}} = {8.33\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{8.33\text{ years}^2}} = {2.9\text{ years}} $ The average porcupine at the zoo is 10 years old. There is a standard deviation of 2.9 years.